Given an array of integers nums, return TRUE if each value in the array has a unique number of occurrences; otherwise, return FALSE.
Constraints:
nums.length <=
nums[i] <=
The algorithm checks if there is a unique number of occurrences of each element by using a hash map and a hash set. First, for each element of the nums array, put the element in the hash map if it does not already exist in the hash map; otherwise, increment its count. Second, put all the elements' frequencies (counts) in a hash set. Finally, the algorithm will return TRUE if the count of the hash map and the hash set is the same and FALSE otherwise.
The algorithm to solve this problem is as follows:
Initialize a hash map and a hash set.
Iterate over the nums array, and for each element:
If the element is not already in the hash map, put the element in the hash map as a key and
Otherwise, if the element already exists in the hash map, increment that element's value by
After the loop terminates, iterate over the hash map and store all ...
Given an array of integers nums, return TRUE if each value in the array has a unique number of occurrences; otherwise, return FALSE.
Constraints:
nums.length <=
nums[i] <=
The algorithm checks if there is a unique number of occurrences of each element by using a hash map and a hash set. First, for each element of the nums array, put the element in the hash map if it does not already exist in the hash map; otherwise, increment its count. Second, put all the elements' frequencies (counts) in a hash set. Finally, the algorithm will return TRUE if the count of the hash map and the hash set is the same and FALSE otherwise.
The algorithm to solve this problem is as follows:
Initialize a hash map and a hash set.
Iterate over the nums array, and for each element:
If the element is not already in the hash map, put the element in the hash map as a key and
Otherwise, if the element already exists in the hash map, increment that element's value by
After the loop terminates, iterate over the hash map and store all ...